NODE ANALYSIS

• One of the systematic ways to determine every voltage and current in a circuit

The variables used to describe the circuit will be “Node Voltages” -- The voltages of each node with respect to a pre-selected reference node

IT IS INSTRUCTIVE TO START THE PRESENTATION WITHA RECAP OF A PROBLEM SOLVED BEFORE USING SERIES/PARALLEL RESISTOR COMBINATIONS

COMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUIT

SECOND: “BACKTRACK” USING KVL, KCL OHM’S

k

VI a

62 :SOHM' 0321 III :KCL

3*3 IkVb :SOHM' …OTHER OPTIONS...

4

34

*4124

12

IkV

II

b

5

345

*3

0

IkV

III

C

:SOHM'

:KCL

k12kk 12||4

k6

kk 6||6

k

VI

12

121

)12(93

3

aV

3I

FIRST REDUCE TO A SINGLE LOOP CIRCUIT

THE NODE ANALYSIS PERSPECTIVETHERE ARE FIVE NODES. IF ONE NODE IS SELECTED AS REFERENCE THEN THERE AREFOUR VOLTAGES WITH RESPECTTO THE REFERENCE NODE

THEOREM: IF ALL NODE VOLTAGES WITHRESPECT TO A COMMON REFERENCE NODEARE KNOWN THEN ONE CAN DETERMINE ANY OTHER ELECTRICAL VARIABLE FORTHE CIRCUIT

aS

aS

VVV

VVV

1

10

ba

ba

VVV

VVV

3

3 0

______caV

QUESTIONDRILL

REFERENCE

SV

aV

bV

cV

1V

KVL

3V

KVL

5V

WHAT IS THE PATTERN???

KVL

05 bc VVV

ONCE THE VOLTAGES ARE KNOWN THE CURRENTS CANBE COMPUTED USING OHM’SLAW

A GENERAL VIEW

Rv

NmR vvv

5 b cV V V

THE REFERENCE DIRECTION FOR CURRENTS IS IRRELEVANT

'i

Rv

USING THE LEFT-RIGHT REFERENCE DIRECTIONTHE VOLTAGE DROP ACROSS THE RESISTOR MUSTHAVE THE POLARITY SHOWN

R

vvi Nm LAW SOHM'

IF THE CURRENT REFERENCE DIRECTION IS REVERSED ...

'Rv

THE PASSIVE SIGN CONVENTION WILL ASSIGNTHE REVERSE REFERENCE POLARITY TO THE VOLTAGE ACROSS THE RESISTOR

R

vvi mN 'LAW SOHM'

'ii PASSIVE SIGN CONVENTION RULES!

DEFINING THE REFERENCE NODE IS VITAL

SMEANINGLES IS4V VSTATEMENT THE 1 UNTIL THE REFERENCE POINT IS DEFINED

BY CONVENTION THE GROUND SYMBOLSPECIFIES THE REFERENCE POINT.

V4

ALL NODE VOLTAGES ARE MEASURED WITHRESPECT TO THAT REFERENCE POINT

V2

_____?12 V

12V

THE STRATEGY FOR NODE ANALYSIS 1. IDENTIFY ALL NODES AND SELECT A REFERENCE NODE

2. IDENTIFY KNOWN NODE VOLTAGES

3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION(e.g.,SUM OF CURRENT LEAVING =0)

0:@321 IIIV

a

4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES

0369

k

VV

k

V

k

VV baasa AND GET ALGEBRAIC EQUATIONS IN THE NODE VOLTAGES ...

REFERENCE

SV

aV

bV

cV

0:@543 IIIV

b

0:@65 IIV

c

0943

k

VV

k

V

k

VVcbbab

039

k

V

k

VVcbc

SHORTCUT:SHORTCUT: SKIP WRITING THESE EQUATIONS...

AND PRACTICE WRITING

THESE DIRECTLY

a b c

d

aV

bV

cV

dV

1R 3R

2R

WHEN WRITING A NODE EQUATION...AT EACH NODE ONE CAN CHOSE ARBITRARYDIRECTIONS FOR THE CURRENTS

AND SELECT ANY FORM OF KCL.WHEN THE CURRENTS ARE REPLACED IN TERMSOF THE NODE VOLTAGES THE NODE EQUATIONSTHAT RESULT ARE THE SAME OR EQUIVALENT

00321

321

R

VV

R

VV

R

VVIII cbdbba

0LEAVING CURRENTS

00321

321

R

VV

R

VV

R

VVIII cbdbba

0 NODE INTO CURRENTS

2I3I1I

a b c

d

aV

bV

cV

dV

1R 3R

2R '2I

'3I'

1I

00321

'3

'2

'1

R

VV

R

VV

R

VVIII bcdbab

0LEAVING CURRENTS

00321

'3

'2

'1

R

VV

R

VV

R

VVIII bcdbab

0 NODE INTO CURRENTS

WHEN WRITING THE NODE EQUATIONSWRITE THE EQUATION DIRECTLY IN TERMSOF THE NODE VOLTAGES.BY DEFAULT USE KCL IN THE FORMSUM-OF-CURRENTS-LEAVING = 0

THE REFERENCE DIRECTION FOR THECURRENTS DOES NOT AFFECT THE NODEEQUATION

CIRCUITS WITH ONLY INDEPENDENT SOURCES

HINT:HINT: THE FORMAL MANIPULATION OF EQUATIONS MAY BE SIMPLER IF ONE USES CONDUCTANCES INSTEAD OF RESISTANCES.

@ NODE 1

02

21

1

1

R

vv

R

viA SRESISTANCE USING

0)( 21211 vvGvGiA ESCONDUCTANC WITH

REORDERING TERMS

@ NODE 2

REORDERING TERMS

THE MODEL FOR THE CIRCUIT IS A SYSTEMOF ALGEBRAIC EQUATIONS

EXAMPLE

@ NODE 1 WE VISUALIZE THE CURRENTSLEAVING AND WRITE THE KCL EQUATION

REPEAT THE PROCESS AT NODE 2

03

12

4

122

R

vv

R

vvi

OR VISUALIZE CURRENTS GOING INTO NODE

WRITE THE KCL EQUATIONS

mA15

A

B

C

k8 k8k2 k2

SELECT ASREFERENCE

AV

BV

MARK THE NODES (TO INSURE THAT NONE IS MISSING)

ANOTHER EXAMPLE OF WRITING NODE EQUATIONS

WRITE KCL AT EACH NODE IN TERMS OFNODE VOLTAGES 015

82@ mA

k

V

k

VA AA

01528

@ mAk

V

k

VB BB

k

VV

k

VmAV

1264:@ 211

1

USING KCL

0126

2:@ 1222

k

VV

k

VmAV

BY “INSPECTION”

mAVk

Vkk

412

1

12

1

6

121

mAVkkk

212

1

6

1

12

12

LEARNING EXTENSION

mA6

1I

2I

3I

036

6:@ 222 k

V

k

VmAV 2

12V V

CURRENTS COULD BE COMPUTED DIRECTLYUSING KCL AND CURRENT DIVIDER!!

1

2

3

8

3(6 ) 2

3 66

(6 ) 43 6

I mA

kI mA mA

k kk

I mA mAk k

LEARNING EXTENSION

IN MOST CASES THEREARE SEVERAL DIFFERENTWAYS OF SOLVING APROBLEM

NODE EQS. BY INSPECTION

mAVVk

6202

121

mAVkk

V 63

1

6

10 21

116V V

k

VI

k

VI

k

VI

3622

32

21

1

Once node voltages are known

1

1@ : 2 6 0

2

VV mA mA

k

Node analysis

3 nodes plus the reference. In principle one needs 3 equations...

…but two nodes are connected tothe reference through voltage sources. Hence those node voltages are known!!!

…Only one KCL is necessary

012126

12322

k

VV

k

VV

k

V

][5.1][64

0)()(2

22

12322

VVVV

VVVVV

EQUATIONS THE SOLVING

Hint: Each voltage sourceconnected to the referencenode saves one node equation

CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES

][6

][12

3

1

VV

VV

THESE ARE THE REMAININGTWO NODE EQUATIONS

We will use this example to introduce the concept of a SUPERNODE

Conventional node analysis requires all currents at a node

@V_1 06

6 1 SIk

VmA

@V_2 012

4 2 k

VmAIS

2 eqs, 3 unknowns...Panic!! The current through the source is notrelated to the voltage of the source

Math solution: add one equation

][621 VVV

Efficient solutionEfficient solution: enclose the source, and all elements in parallel, inside a surface.

Apply KCL to the surface!!!

04126

6 21 mAk

V

k

VmA

The source current is interiorto the surface and is not required

We STILL need one more equation

][621 VVV Only 2 eqs in two unknowns!!!

SUPERNODE

THE SUPERNODE TECHNIQUE

SI

k 12 / *

][6

046126

21

21

VVV

mAmAk

V

k

V

(2)

(1)

Equations The

ALGEBRAIC DETAILS

...by Multiply Eq(1). in rsdenominato Eliminate 1.

Solution

][6

][242

21

21

VVV

VVV

][10][303 11 VVVV 2 Veliminate to equations Add2.

][4][612 VVVV 2 Vcompute to Eq(2) Use 3.

VV

VV

4

6

4

1

SOURCES CONNECTED TO THE

REFERENCE

SUPERNODE

CONSTRAINT EQUATION VVV 1223

KCL @ SUPERNODE

02

)4(

212

63322

k

V

k

V

k

V

k

Vk2/*

VVV 22332VVV 12

32 add and 3/*

VV 385 3

mAk

VIO

8.323 LAW SOHM'

OIV FOR NEEDED NOT IS

2